LeetCode - 2.Add Two Numbers

原创文章转载请注明出处

Add Two Numbers
You are given two linked lists representing two non-negative numbers. The digits are stored in reverse order and each of their nodes contain a single digit. Add the two numbers and return it as a linked list.
Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)
Output: 7 -> 0 -> 8

其实就是列加法竖式，C的速度无人能敌。我所熟悉的语言中，JavaScript依然垫底，但是同样是编译型的Swift会差Go那么多实在是大跌眼镜，指针和结构体初始化的操作效率还是高。

Paste_Image.png

Swift

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     public var val: Int
 *     public var next: ListNode?
 *     public init(_ val: Int) {
 *         self.val = val
 *         self.next = nil
 *     }
 * }
 */
class Solution {
    func addTwoNumbers(_ l1: ListNode?, _ l2: ListNode?) -> ListNode? {
        var currentL1 = l1
        var currentL2 = l2
        let rootNode = ListNode(0)
        var currentNode : ListNode? = rootNode
        var overflow : Int = 0
        
        while (currentL1 != nil || currentL2 != nil || overflow != 0) {
            
            // compute current node
            let sum = overflow + (currentL1?.val ?? 0) + (currentL2?.val ?? 0)
            if sum >= 10 {
                currentNode?.next = ListNode( sum - 10 ) 
                overflow = 1
            } else {
                currentNode?.next = ListNode( sum )
                overflow = 0
            }
            
            // prepare data for next step
            currentNode = currentNode?.next
            currentL1 = currentL1?.next
            currentL2 = currentL2?.next
        }
        
        return rootNode.next
    }
}

Go

/**
 * Definition for singly-linked list.
 * type ListNode struct {
 *     Val int
 *     Next *ListNode
 * }
 */
func addTwoNumbers(l1 *ListNode, l2 *ListNode) *ListNode {
    overflow := 0
    head := new(ListNode)
    cur := head
    for l1 != nil || l2 != nil || overflow != 0 {
        n1, n2 := 0, 0
        if l1 != nil {
            n1, l1 = l1.Val, l1.Next
        }
        if l2 != nil {
            n2, l2 = l2.Val, l2.Next
        }
        num := n1 + n2 + overflow
        if num >= 10 {
            overflow = 1
            cur.Next = &ListNode{Val:num-10, Next:nil}    
        } else {
            overflow = 0
            cur.Next = &ListNode{Val:num, Next:nil}
        }
        cur = cur.Next
    }
    return head.Next   
}

我是咕咕鸡，一个还在不停学习的全栈工程师。
热爱生活，喜欢跑步，家庭是我不断向前进步的动力。